What is the formula for the sequence #1, 3, 7, 9,...# ?

2 Answers
Sep 20, 2017

#n+(n-1)#

(assuming that the sequence ie 1,3,5,7,9)

Explanation:

#"n of " 2=3#
#"n of " 3=5#
#"n of " 4=7#
#"n of " 5=9#

Basically you just work out how different #a# is from #n# and adjust to suit. In this case, the pattern is the #n# plus what the previous #n# was: #n-1#.

Checking my idea of the pattern:
If #n = 1#, then a = #1+(1-1)=1+0=1# which is correct.

If #n = 7#, then a = #5+(5-1) = 5 + 4 = 9# which is also correct.

If #n = k#, then a = #k +(k-1) = 2k-1# which when k is substituted in is the same as above #(2*5=10)#

Sep 20, 2017

This sequence can be matched by:

#a_n = 3n-5/2-(-1)^n/2#

or by:

#a_n = -2/3n^3+5n^2-25/3n+5#

Explanation:

Assuming that the question is correct as given, there are several way to match the sequence:

#1, 3, 7, 9#

with a formula.

For example, if we add #(-1)^n/2# to each element, then we get the sequence:

#1/2, 7/2, 13/2, 19/2#

which is an arithmetic sequence with common difference #3#.

Hence we can write a formula for the given sequence:

#a_n = 3n-5/2-(-1)^n/2#

Alternatively, we could use the method of differences to match the sequence with a polynomial...

Write down the given sequence:

#color(blue)(1), 3, 7, 9#

Write down the sequence of differences between each consecutive pair of terms:

#color(purple)(2), 4, 2#

Write down the sequence of differences of those differences:

#color(brown)(2), -2#

Write down the sequence of differences of those differences:

#color(green)(-4)#

Having arrived at a constant sequence (albeit of just one element), we can use the initial term of each of these sequences as coefficients to give us a direct formula:

#a_n = color(blue)(1)/(0!) + color(purple)(2)/(1!)(n-1) + color(brown)(2)/(2!)(n-1)(n-2) + color(green)(-4)/(3!)(n-1)(n-2)(n-3)#

#color(white)(a_n) = 1+2n-2+n^2-3n+2-2/3n^3+4n^2-22/3n+4#

#color(white)(a_n) = -2/3n^3+5n^2-25/3n+5#